Get the Voltage from the I(A)-Time diagram

  • Thread starter Thread starter papm
  • Start date Start date
  • Tags Tags
    Diagram Voltage
AI Thread Summary
The area under the I(A)-Time diagram represents the capacity, calculated as Q=I(A)*t(s). To determine the voltage for energy calculations of a battery and capacitor, additional information is necessary, as current and time alone do not provide voltage values. The discussion highlights that the voltage drop across the battery is not equal to I*R unless in a short-circuit scenario, and the actual energy delivered can be calculated using the integral of voltage and current over time. For practical applications, the rated voltage of the battery should be used to compute the delivered energy. Accurate voltage measurements or references are essential for determining the energy output of the system.
papm
Messages
4
Reaction score
0
Thread moved from the technical forums, so no Template is shown
244392

The area under the diagramm is the capacity.Q=I(A)*t(s)
How can I can get the Voltage so tha I get the energy of the battery and capacitor?
 
Physics news on Phys.org
Welcome to PF.
I assume you are talking about the discharge of a chemical battery.
papm said:
How can I can get the Voltage so tha I get the energy of the battery and capacitor?
Because only current and time are specified, it is not possible to know the voltage of the battery.
Does the word "capacitor" you use actually refer to the capacity of the battery in kW*hours, or do you have a capacitor in the circuit.
 
Thank you for the reply
I have a chemical battery and a capacitor (over 45(A)).
Can I find the energy of them?
Can I assume V=I*R with R=1ohm?
Because I(A) get straight to 45(A) and we don't have a curve.
 
papm said:
Can I assume V=I*R with R=1ohm?
Where does the idea of 1 ohm come from, why not 1 milliohm, or 1k ?
 
Baluncore said:
Where does the idea of 1 ohm come from, why not 1 milliohm, or 1k ?
Don't know I just assumed to find out the energy.
Maybe becouse we get the I=45(A) immediately?
 
We need more information.
You need to find the voltage from other references or make a measurement.
 
This exercise was on last exams.
The only information was the diagramm that is from a electric car with a battery and and a capacitor for acceleration. There is no more information.
 
That diagram will contain the detail.
 
  • Like
Likes papm
papm said:
Because I(A) get straight to 45(A) and we don't have a curve.
That's 50A isn't it? That graph can tell you the net charge into/out of the capacitor. Then use Q = C.V to determine the associated change in voltage, but you do need to know C.

There'll be a graph describing the battery, and you use that to determine the battery voltage.
 
  • #10
This is cycle of a functioning electric vehicle [as translated from Greek]
Let's say this is total discharge battery= 117.6 Ah .A usual battery of 12V 120 Ah could present this discharge process. However a 24 V battery could present the same.
It could be one like this:12V/120Ah/115F51 JIS Car batteries, Sealed lead acid batteries -in my opinion.
 
  • #12
By the way, I*R it is not the battery voltage-except in short-circuit case-but it is the voltage drop through the battery and so I^2*R it is not the delivery power but the battery losses.
The delivery energy it is ʃV*I*dt .Since in the discharge time interval [up to- let's say 1.8 V from 2.2 V rated] you may consider the delivered voltage as constant and equal to the rated and then the delivered energy will be Vrated*Q.
 

Similar threads

Replies
3
Views
2K
Replies
4
Views
2K
Replies
10
Views
3K
Replies
15
Views
2K
Replies
6
Views
2K
Back
Top